Proposition 5.13 (ii) in Scholze's Perfectoid Spaces

Question

In Proposition 5.13 (ii) in Scholze's Perfectoid Spaces, we have $R \to S$ a morphism of $\Bbb F_p$-algebras and the assumption that the relative Frobenius $\Phi_{S/R}$ induces an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} S \to S_{(\Phi)}$ in $D(R)$, the derived category of $R$-modules. Here $R_{(\Phi)}$ is the ring $R$ with the $R$-algebra structure given by the Frobenius $R \to R$, and $S_{(\Phi)}$ is similarly defined.

Then they go on to claim, in the proof of (ii), that this assumption says that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ induces a quasi-isomorphism of simplicial algebras, where $S^\bullet$ is a simplicial resolution of $S$ by free $R$-algebras. I do not understand why. My guess is that this is because the complex $R_{(\Phi)} \otimes_R^{\Bbb L} S \in D(R)$ in the assumption is constructed by taking such a resolution of $S$, so $R_{(\Phi)} \otimes_R^{\Bbb L} S \triangleq R_{(\Phi)} \otimes_R S^\bullet \in D(R)$, and then $S^\bullet_{(\Phi)} \to S_{(\Phi)}$ is a quasi-isomorphism because $S^\bullet_{(\Phi)}$ is a resolution of $S_{(\Phi)}$, so in $D(R)$ we have $R_{(\Phi)} \otimes_R S^\bullet \triangleq R_{(\Phi)} \otimes_R^{\Bbb L} S \cong S_{(\Phi)} \cong S^\bullet_{(\Phi)}$?

After that, they say that this implies that $\Phi_{S^\bullet/R}: R_{(\Phi)} \otimes_R S^\bullet \to S^\bullet_{(\Phi)}$ gives an isomorphism $R_{(\Phi)} \otimes_R^{\Bbb L} \Bbb L_{S/R} \cong \Bbb L_{S_{(\Phi)}/R_{(\Phi)}}$. I also do not understand why. According to the same paper, the complex $\Bbb L_{S/R}$ is defined to be $\Omega^{1}_{S_\bullet/R} \otimes_{S_\bullet} S$, but I do not see a relation. The reference for this part is Lemma 6.5.9 of Gabber–Romero, which references Proposition II.1.2.6.2 of Illusie's Complexe Cotangent et Déformations I, but I do not see how to apply the proposition to this situation.

Finally, both Scholze and Gabber–Romero claim that $\Bbb L_{S_{(\Phi)}/R_{(\Phi)}} \cong \Bbb L_{S/R}$, but I do not know why. I think it is because $R_{(\Phi)}$ is defined to be $R$ as a ring, and the map $R_{(\Phi)} \to S_{(\Phi)}$ as rings should be the same as the map $R \to S$ as rings, so their modules of Kähler differentials should be the same.

P.S. I don't understand why Scholze uses $S_\bullet$ in the definition of $\Bbb L_{S/R}$ when he uses $S^\bullet$. Indeed, Gabber–Romero uses $P^\bullet$ in Lemma 6.5.9.

Answer 1

Regarding the first and third question, what you say is correct. For the second question, you are looking for the base change compatibility of the cotangent complex: If $R\to R'$ is any map of rings and $S$ is an $R$-algebra such that $S'=S\otimes^L_R R'$ sits in degree $0$, then $$R'\otimes^L_R \mathbb L_{S/R}\cong \mathbb L_{S'/R'}.$$

To see this, use any simplicial resolution $S_\bullet$ of $S$ by free $R$-algebras; then $S_\bullet\otimes_R R'$ is such a resolution of $S'$. The left-hand side is computed by the simplicial $R'$-module $R'\otimes_R(\Omega^1_{S_\bullet/R}\otimes_{S_\bullet} S)$, and the right-hand side by $\Omega^1_{S'_\bullet/R'}\otimes_{S'_\bullet} S'$; these two simplicial $R'$-modules agree.

Implicit here is that one can use any free resolution to compute the cotangent complex, not necessarily the standard functorial one.

But I think there must be some more basic confusion on cotangent complexes, as you say that you do not see the relation between $\mathbb L_{S/R}$ and $\Omega^1_{S_\bullet/R}\otimes_{S_\bullet} S$. To address this: What do you take as the definition of the cotangent complex?